# Problem 2 – Hint 5

Let us sum up the previous hints, and make them clearer here and there.

The process starts with homogeneous heating: the entire wire has (almost) the same temperature which is increasing in time. This continues until the critical temperature is reached. At this moment, an instability develops: from the hottest point on the wire (which is the hottest due to small fluctuations along the wire), a phase transition boundary spreads along the wire until a thermal equilibrium is reached. At this new thermal equilibrium, high- and low-resistivity wire segments coexist, hence the integral resistance of the wire is now bigger than it was before the instability developed. Since the thermalization time is assumed to be fast as compared to the time scale at which the voltage is changed, the increase of the integral resistivity during the evolution of the instability could be seen as a jump, if we were to plot the net resistance alongside with the total power. Note that a point $Q$ on the wire separating the high- and low-resistivity phases must be at the critical temperature (since to one side from $Q$, $T while to the other side $T>T_c$). At the immediate neighbourhood of $Q$, thermal flux along the wire cannot be neglected and that’s where the temperature depends on the coordinate $x$ along the wire. According to the assumptions of the problem, this temperature transition range is narrow; let $\delta$ denote its width. Then, at $x, the temperature of the wire is almost constant and equal to $T_L$, and at $x>x_Q+\delta$, the temperature of the wire is almost constant and equal to $T_R$; here, $x_Q$ denotes the coordinate of the phase separation point $Q$, and the values of $T_L$ and $T_R$ are to be found from the thermal equilibrium conditions. There are two conditions to consider: first, there is no net thermal flux along the wire at the point $Q$ (because otherwise, the position of the point $Q$ would be pushed according to the direction of the net thermal flux; note also that we know the temperature of the point $Q$!); second, the power dissipation at $x>x_Q+\delta$ and at $x differ by a factor of two (because it is related to $I^2\rho$, where $I$ is the current strength in the wire). An important consequence is that as long as a phase separation points exist, the temperatures of the different regions on the wire remain constant in time and equal to $T_L$ and $T_R$, respectively; what is changing is the width of these regions. Since the temperature of the wire for a fixed phase is defined by the current strength $I$, the current strength $I$ must also remain constant – as long as different phases coexist.

There will be no more hints for the Problem 2! The next intermediate results for the Problem 2 will be published at 13:00 GMT, 5th February 2023.

Please submit the solution to this problem via e-mail to physcs.cup@gmail.com.